InterviewSolution
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InterviewSolution
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A ring of radius R having unifromly distributed charge Q is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is `T_0.` Now a vertical magnetic field is switched on and ring is rotated at constant angular velocity `omega`. Find the maximum `omega` with which the ring can be rotated if the strings can withstand a maximum tension of `3T_0 //2.` |
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Answer» In equilibrium, `2T_0=mg` or `T_0=(mg)/2….(i)` magnetic moment, `M=iA=(omega/(2pi)Q)(piR^2)` `tau=MB sin90^@=(omegaBQR^2)/2` Let `T_1 and T_2` be the tensions in the two strings when magnetic field is switched on `(T_1gtT_2)` For translation equilibrium, `T_1+T_2=mg....(ii)` For rotational equilibrium `(T_1-T_2)D/2=tau=(omegaBQR^2)/2` or `T_1-T_2=(omegaBQR^2)/D....(iii)` Solving Eqs. (ii) and (iii) we have `T_1=(mg)/2+(omegaBQR^2)/(2D)` As `T_1gtT_2` and maximum values of `T_1` can be `3T_0//2`, we have `(3T_0)/2=T_0+(omega_(max)BQR^2)/(2D) ((mg)/2=T_0)` `:. omega_(max)=(DT_0)/(BQR^2)` |
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